Problem: Find $\dfrac{d}{dx}\left(\dfrac{e^x}{\sqrt{x}}\right)$. Choose 1 answer: Choose 1 answer: (Choice A) A $e^x-\dfrac{1}{2\sqrt{x}}$ (Choice B) B $2\sqrt{x}e^x$ (Choice C) C $\dfrac{e^x\left(\sqrt{x}-\dfrac{1}{2\sqrt{x}}\right)}{x}$ (Choice D) D $e^x\left(\sqrt{x}-\dfrac{1}{2\sqrt{x}}\right)$
Explanation: $\dfrac{e^x}{\sqrt{x}}$ is the quotient of two, more basic, expressions: $e^x$ and $\sqrt{x}$. Therefore, the derivative of the expression can be found using the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left(\dfrac{e^x}{\sqrt{x}}\right) \\\\ &=\dfrac{\dfrac{d}{dx}(e^x)\sqrt{x}-e^x\dfrac{d}{dx}(\sqrt{x})}{(\sqrt{x})^2}&&\gray{\text{The quotient rule}} \\\\ &=\dfrac{e^x\cdot \sqrt{x}-e^x\cdot \dfrac1{2\sqrt{x}}}{(\sqrt{x})^2}&&\gray{\text{Differentiate }e^x\text{ and }\sqrt{x}} \\\\ &=\dfrac{e^x\left(\sqrt{x}-\dfrac{1}{2\sqrt{x}}\right)}{x}&&\gray{\text{Simplify}} \end{aligned}$ In conclusion, $\dfrac{d}{dx}\left(\dfrac{e^x}{\sqrt{x}}\right)=\dfrac{e^x\left(\sqrt{x}-\dfrac{1}{2\sqrt{x}}\right)}{x}$